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(A) $1$. The reaction of $CH_2(O)CH-CH_2-Br$ with $Mg$ in dry ether forms a Grignard reagent $(A)$,which is $CH_2(O)CH-CH_2-MgBr$.$2$. This Grignard r

Vedclass · Accepted Answer

$CH_2=CH-CH_2-O-CH_3$

Vedclass · Answer

(A) $1$. The reaction of $CH_2(O)CH-CH_2-Br$ with $Mg$ in dry ether forms a Grignard reagent $(A)$,which is $CH_2(O)CH-CH_2-MgBr$.$2$. This Grignard reagent undergoes an intramolecular acid-base reaction where the carbanion attacks the epoxide ring,leading to ring opening and the formation of an alkoxide intermediate: $CH_2=CH-CH_2-OMgBr$.$3$. This alkoxide then acts as a nucleophile and attacks $CH_3-I$ in an $S_N2$ reaction to form the final product $(B)$,which is $CH_2=CH-CH_2-O-CH_3$.

Product $(B)$ is

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